Question: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x+4y &= 2 \\ -2x+3y &= -4\end{align*}$
Solution: Begin by moving the $x$ -term in the second equation to the right side of the equation. $3y = 2x-4$ Divide both sides by $3$ to isolate $y$ $y = {\dfrac{2}{3}x - \dfrac{4}{3}}$ Substitute this expression for $y$ in the first equation. $-8x+4({\dfrac{2}{3}x - \dfrac{4}{3}}) = 2$ $-8x + \dfrac{8}{3}x - \dfrac{16}{3} = 2$ Simplify by combining terms, then solve for $x$ $-\dfrac{16}{3}x - \dfrac{16}{3} = 2$ $-\dfrac{16}{3}x = \dfrac{22}{3}$ $x = -\dfrac{11}{8}$ Substitute $-\dfrac{11}{8}$ for $x$ back into the top equation. $-8( -\dfrac{11}{8})+4y = 2$ $11+4y = 2$ $4y = -9$ $y = -\dfrac{9}{4}$ The solution is $\enspace x = -\dfrac{11}{8}, \enspace y = -\dfrac{9}{4}$.